$H_0^1(Omega)hookrightarrow L^4(Omega)$?What is the Sobolev Lemma?Differences between $-Delta: H_0^1(Omega)to H^-1(Omega)$ and $-Delta: H^2(Omega)cap H_0^1(Omega)to L^2(Omega)$functions in $H_0^1(Omega) cap C(overlineOmega)$ are zero on the boundaryReference request: The compactness and compact embedding in Besov Space?$u in W^2,2(Omega)$ implies $Delta_p u in L^q(Omega)$ for some $q>1$?Problem to understand Sobolev imbedding.Inequality in $W^1,infty(Omega)$Show that $ C^1(overlineOmega) hookrightarrow C^0,alpha(overlineOmega)$ is compact.Compact embedding of $W^k,p(Omega)$ into $W^k-1,p*(Omega)$, where $p^*$ is the Sobolev conjugateEquivalence of norms in the space $H_Delta(Omega)$
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$H_0^1(Omega)hookrightarrow L^4(Omega)$?
What is the Sobolev Lemma?Differences between $-Delta: H_0^1(Omega)to H^-1(Omega)$ and $-Delta: H^2(Omega)cap H_0^1(Omega)to L^2(Omega)$functions in $H_0^1(Omega) cap C(overlineOmega)$ are zero on the boundaryReference request: The compactness and compact embedding in Besov Space?$u in W^2,2(Omega)$ implies $Delta_p u in L^q(Omega)$ for some $q>1$?Problem to understand Sobolev imbedding.Inequality in $W^1,infty(Omega)$Show that $ C^1(overlineOmega) hookrightarrow C^0,alpha(overlineOmega)$ is compact.Compact embedding of $W^k,p(Omega)$ into $W^k-1,p*(Omega)$, where $p^*$ is the Sobolev conjugateEquivalence of norms in the space $H_Delta(Omega)$
$begingroup$
In a paper I see that the authors used $H_0^1(Omega)hookrightarrow L^4(Omega)$ where $Omega$ is an open bounded domain in $mathbbR^N$ with smooth boundary. I think that this imbedding holds only for those $N$ which satisfy $4leq 2^star$($2^star=frac2NN-2$) i.e. when $Omega$ is in $mathbbR^3$ or $mathbbR^4$. How we can show this imbedding for higher dimensions?
real-analysis functional-analysis analysis pde sobolev-spaces
asked Mar 16 at 9:49
AlbertAlbert
739312
$endgroup$
add a comment |
$begingroup$
In a paper I see that the authors used $H_0^1(Omega)hookrightarrow L^4(Omega)$ where $Omega$ is an open bounded domain in $mathbbR^N$ with smooth boundary. I think that this imbedding holds only for those $N$ which satisfy $4leq 2^star$($2^star=frac2NN-2$) i.e. when $Omega$ is in $mathbbR^3$ or $mathbbR^4$. How we can show this imbedding for higher dimensions?
real-analysis functional-analysis analysis pde sobolev-spaces
asked Mar 16 at 9:49
AlbertAlbert
739312
$endgroup$
1
$begingroup$
You are right, this is only true for $Nle 4$.
$endgroup$
– daw
Mar 16 at 12:00
$begingroup$
@daw Thank you very much. I do think so.
$endgroup$
– Albert
Mar 16 at 17:11
add a comment |
$begingroup$
In a paper I see that the authors used $H_0^1(Omega)hookrightarrow L^4(Omega)$ where $Omega$ is an open bounded domain in $mathbbR^N$ with smooth boundary. I think that this imbedding holds only for those $N$ which satisfy $4leq 2^star$($2^star=frac2NN-2$) i.e. when $Omega$ is in $mathbbR^3$ or $mathbbR^4$. How we can show this imbedding for higher dimensions?
real-analysis functional-analysis analysis pde sobolev-spaces
asked Mar 16 at 9:49
AlbertAlbert
739312
$endgroup$
In a paper I see that the authors used $H_0^1(Omega)hookrightarrow L^4(Omega)$ where $Omega$ is an open bounded domain in $mathbbR^N$ with smooth boundary. I think that this imbedding holds only for those $N$ which satisfy $4leq 2^star$($2^star=frac2NN-2$) i.e. when $Omega$ is in $mathbbR^3$ or $mathbbR^4$. How we can show this imbedding for higher dimensions?
real-analysis functional-analysis analysis pde sobolev-spaces
real-analysis functional-analysis analysis pde sobolev-spaces
asked Mar 16 at 9:49
AlbertAlbert
739312
asked Mar 16 at 9:49
AlbertAlbert
739312
edited Mar 17 at 6:18
Albert
asked Mar 16 at 9:49
AlbertAlbert
739312
asked Mar 16 at 9:49
AlbertAlbert
739312
asked Mar 16 at 9:49
AlbertAlbert
739312
739312
1
$begingroup$
You are right, this is only true for $Nle 4$.
$endgroup$
– daw
Mar 16 at 12:00
$begingroup$
@daw Thank you very much. I do think so.
$endgroup$
– Albert
Mar 16 at 17:11
add a comment |
1
$begingroup$
You are right, this is only true for $Nle 4$.
$endgroup$
– daw
Mar 16 at 12:00
$begingroup$
@daw Thank you very much. I do think so.
$endgroup$
– Albert
Mar 16 at 17:11
1
1
$begingroup$
You are right, this is only true for $Nle 4$.
$endgroup$
– daw
Mar 16 at 12:00
$begingroup$
You are right, this is only true for $Nle 4$.
$endgroup$
– daw
Mar 16 at 12:00
$begingroup$
@daw Thank you very much. I do think so.
$endgroup$
– Albert
Mar 16 at 17:11
$begingroup$
@daw Thank you very much. I do think so.
$endgroup$
– Albert
Mar 16 at 17:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The general result is called the Sobolev embedding theorem, which states that $H_0^k$ is embedded in $L^q$ where $$frac1q=frac12-frackn$$ where $n$ is the dimension of the space. This can be generalized even further, see the Wikipedia article.
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
$endgroup$
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
The general result is called the Sobolev embedding theorem, which states that $H_0^k$ is embedded in $L^q$ where $$frac1q=frac12-frackn$$ where $n$ is the dimension of the space. This can be generalized even further, see the Wikipedia article.
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
$endgroup$
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
add a comment |
$begingroup$
The general result is called the Sobolev embedding theorem, which states that $H_0^k$ is embedded in $L^q$ where $$frac1q=frac12-frackn$$ where $n$ is the dimension of the space. This can be generalized even further, see the Wikipedia article.
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
$endgroup$
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
add a comment |
$begingroup$
The general result is called the Sobolev embedding theorem, which states that $H_0^k$ is embedded in $L^q$ where $$frac1q=frac12-frackn$$ where $n$ is the dimension of the space. This can be generalized even further, see the Wikipedia article.
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
$endgroup$
The general result is called the Sobolev embedding theorem, which states that $H_0^k$ is embedded in $L^q$ where $$frac1q=frac12-frackn$$ where $n$ is the dimension of the space. This can be generalized even further, see the Wikipedia article.
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
answered Mar 16 at 11:41
ChrystomathChrystomath
1,783513
1,783513
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
add a comment |
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
$begingroup$
Thank you very much for your answer. I know the generalized form of the Sobolev imbedding. But my question is in this special case $k=1$.
$endgroup$
– Albert
Mar 16 at 17:16
add a comment |
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$begingroup$
You are right, this is only true for $Nle 4$.
$endgroup$
– daw
Mar 16 at 12:00
$begingroup$
@daw Thank you very much. I do think so.
$endgroup$
– Albert
Mar 16 at 17:11